A small metal sphere of mass 40 mg and density 8000 kg m ⁻³ is dropped into a tall column of an unknown liquid. After some time, it attains a constant velocity. If the viscous force acting on the sphere at this stage is 3 10⁻⁴ N , the density of the liquid is x 10³ kg m ⁻³ . The value of x is _____ . [Use g = 10 m s ⁻² ]

Step-by-step solution

When the sphere attains a constant velocity, the net force acting on it is zero. Therefore, the sum of the viscous force and the buoyant force equals the weight of the sphere. F_ v + F_ B = W The weight of the sphere is: W = mg = 40 10⁻⁶ kg 10 m s ⁻² = 4 10⁻⁴ N The buoyant force is: F_ B = W - F_ v = 4 10⁻⁴ N - 3 10⁻⁴ N = 10⁻⁴ N The volume of the sphere is: V = m _ s = 40 10⁻⁶ 8000 = 5 10⁻⁹ m ³ The buoyant force can also be written as: F_ B = V _ L g Substituting the values: 10⁻⁴ = 5 10⁻⁹ _ L 10 _ L = 10⁻⁴ 5 10⁻⁸ =